rest energy question


by asdf1
Tags: energy, rest
asdf1
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#1
Aug17-05, 02:43 AM
P: 741
Why is the rest energy mc^2 of an electron or positron 0.51 MeV?
Because if it's not moving, why is there "c"?
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inha
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#2
Aug17-05, 10:17 AM
P: 576
E=mc^2. Plug in the electrons mass and divide my c^2. You get m=0.511 MeV/c^2.
c^2 is a constant of propotionality in this case and nothing more.
Zelos
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#3
Aug18-05, 02:24 AM
P: 76
its just the constant of the energy matter. no mroe it dont have to do with the speed.

asdf1
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#4
Aug18-05, 03:26 AM
P: 741

rest energy question


isn't c= the speed of light =3.0*10^8?
Igor_S
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#5
Aug18-05, 05:45 AM
P: 96
Quote Quote by asdf1
isn't c= the speed of light =3.0*10^8?
Yes, speed of light. Not speed of electrons. Think of it just as a constant with which you have to multiply mass to find energy (if it's easier that way)...
asdf1
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#6
Aug18-05, 10:06 AM
P: 741
hmm... but i still think it's weird that you have to taked into account the speed of light a "rest" energy formula, because the electron's not moving...
Igor_S
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#7
Aug19-05, 08:54 AM
P: 96
You may find this link interesting:

http://www.pbs.org/wgbh/nova/einstein/

It explains stuff about the equation you were asking. ;)
Zelos
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#8
Aug19-05, 10:47 AM
P: 76
if u travel at 3*10^8m/s uŽll travel at ftl. the exact value of c is 299792458 m/s.
asdf1
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#9
Aug20-05, 11:54 PM
P: 741
thanks! :)
pervect
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#10
Aug21-05, 02:14 PM
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Quote Quote by asdf1
Why is the rest energy mc^2 of an electron or positron 0.51 MeV?
Because if it's not moving, why is there "c"?
There is an important relationship between the energy and momentum of the electron

E^2 - (pc)^2 = (mc^2)^2

This is a very valuable and useful relationship, but it requires that when p=0 that E is non-zero. We can substitute p=0 in the above equation and find

E=mc^2

While we could write (E' + mc^2) - (pc)^2 = (mc^2)^2, where E' is the kinetic energy of the electron (excluding it's rest energy) it is much more convenient to make E include the rest energy of the electron. So that is what is done.

The total energy of the electron (kinetic plus rest) also plays an important role in other aspects of physics, making the convention E=mc^2 when p=0 very useful, so useful that it is now a standard part of the definitions of physics.
pmb_phy
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#11
Aug21-05, 02:57 PM
P: 2,955
Quote Quote by asdf1
Why is the rest energy mc^2 of an electron or positron 0.51 MeV?
Because if it's not moving, why is there "c"?
It appears clear to me that you're asking why the rest energy is 0.51MeV. If so then I doubt anyone knows why. If there is then it lies outside of classical mechanics and dips into particle theory. As such I'd ask Reilly this question. He's the expert on that subject here.

Pete
selfAdjoint
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#12
Aug21-05, 05:14 PM
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Quote Quote by pmb_phy
It appears clear to me that you're asking why the rest energy is 0.51MeV. If so then I doubt anyone knows why. If there is then it lies outside of classical mechanics and dips into particle theory. As such I'd ask Reilly this question. He's the expert on that subject here.

Pete
No he's plainly asking why this rest energy is specified in tables as [tex]mc^2[/tex]. And it's because in it's rest frame the electron, like any massive particle, has no other energy than it's mass equivalence energy, which as explained earlier in this thread is [tex]mc^2[/tex]. Although the electron is not moving in its rest frame, the basic role of c in the theory of relativity means it appears in this formula.
pervect
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#13
Aug21-05, 11:00 PM
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I was thinking about my earlier response, and it''s fine as far as it goes but it's incomplete.

Probably the clearest physical experiment which demonstrates E=mc^2 is the creation of an electron-positron pair from gamma radiation, or the inverse process of the creation of gamma radiation from electron-positron annhiliaton.

You find that the kinetic energy of the gamma rays generated (by annhilation) or needed (for pair creation) is E=mc^2 as a minimum. The minimum is achieved when the annhilating electron-positron pair (or the created pair) is at rest.

It's also true as I remarked previously that E^2 = (mc^2)^2+(pc)^2 when E is defined as the total energy of the electron, i.e. E=(mc^2)+(kinetic energy), p is the momentum of the electron, and m is the rest mass (aka invariant mass) of the electron.
pmb_phy
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#14
Aug22-05, 02:49 AM
P: 2,955
Quote Quote by selfAdjoint
No he's plainly asking why this rest energy is specified in tables as [tex]mc^2[/tex]. ....
I seriously doubt that but only asdf1 knows for sure. Take a look at the question again

"Why is the rest energy mc^2 of an electron or positron 0.51 MeV?"

This question is easily rephrased by removing the clarification "mc^2" to give

"Why is the rest energy an electron 0.51 MeV?"

Classical relativity can't give that info. I don't see that the question was addressed at all here.

The answer to his question about "c" is most easily given by providing a derivation and let them watch the "c" pop out. See

http://www.geocities.com/physics_wor...ergy_equiv.htm

But only asdf1 can tell us what is "plainly" the question.

Pete
asdf1
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#15
Aug22-05, 04:13 AM
P: 741
@@a
i was wondering why rest energy has "c" in it's formula~
because "c" is usually referred to as the speed of light, so i thought that it was weird to have something that is moving included in a "rest" energy formula~
pmb_phy
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#16
Aug22-05, 06:46 AM
P: 2,955
Quote Quote by asdf1
@@a
i was wondering why rest energy has "c" in it's formula~
because "c" is usually referred to as the speed of light, so i thought that it was weird to have something that is moving included in a "rest" energy formula~
Then I'd say follow the derivation and you'll see that the "c" has nothing to do with the velocity of the particle. This is not an expression likek kinetic energy where the velocity is that of the particle.

Pete
El Hombre Invisible
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#17
Aug22-05, 08:09 AM
P: 1,017
Quote Quote by asdf1
@@a
i was wondering why rest energy has "c" in it's formula~
because "c" is usually referred to as the speed of light, so i thought that it was weird to have something that is moving included in a "rest" energy formula~
asdf1 (lazy username, btw): one of Einstein's postulates providing the basis of special relativity states that the laws of nature must hold in any inertial frame of reference. Once he formulated SR, it became clear that many so-called laws did not fit the bill, so his next job was to rewrite these classical laws so that they did apply in all frames. You'll see 'c' popping up in all of these laws, since 'c' has a special significance in SR: it is required to translate co-ordinates and values from one frame to another, regardless of the velocity of these frames or the observables within. The reason why 'c' is so important is due to the other postulate of SR: that the speed of light in a vacuum is the same in all inertial frames. This speed limit is worked into relativistic mechanics by means of the Lorentz transformation, which gives us the rules of converting the co-ordinates of one frame to another.
asdf1
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#18
Aug22-05, 10:18 AM
P: 741
thanks! :)
lol
the username asdf1 rocks because it's easy to type...


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